I've been trying to find an answer to this question, but have really been stumped so far.

The work-energy principle says that work done on a single particle is equal to its change in kinetic energy. Now let's say a particle is moving in the +x direction at constant speed v and we perform work to reverse its direction so that it moves in the -x direction at constant speed v. This clearly requires work, but its change in kinetic energy is zero, because it has the same speed at the beginning and the end.

Sorry if there is an obvious answer to this, but have been bricking my head until now!

  • $\begingroup$ suppose one throws a ball up with velocity say v0...the gravity does work and at height h the velocity=0, again it reverses its path due to work being done by the same forces and it acquires a velocity v0 at the surface of earth....something is happening..work was being done..work is a scalar product of two vectors ,the force and displacement...it may result in positive and negative physical work and the sum may be zero...it may change the total energy or forms of energy..well I just tried to think aloud.pl. try to find... $\endgroup$ – drvrm Aug 25 '18 at 23:00
  • $\begingroup$ As mentioned below, negative work is done to bring the object to rest, and and equal amount of positive work is done to bring the object back to its original speed. From a physics standpoint, the net work is zero. However, in the real world, assume that you used a gasoline powered device to do this work. You had to pay for the gasoline to stop the object, and you had to pay for an equal amount of gasoline to get the object back up to speed. In my opinion, your misunderstanding stems from the subtlety in the physics definition of net work. $\endgroup$ – David White Aug 26 '18 at 2:08

The answer is that it doesn't actually require work. Using the equation $dW=\vec{F}\cdot d \vec{x}$, we can see that the work done on the object is positive if the force is pushing in the same direction as the object is moving, (speeding the object up) while it is negative if the force points in the opposite direction (slowing the object down). In the case of reversing an object's velocity, negative work is done on the object to slow it down to a state of motionlessness, and this is exactly canceled by the positive work done to speed it back up in the other direction.

If you want a picture for this, imagine a ball rolling partway up a steep hill, and then rolling back the way it came. The ball's velocity has been reversed, but the hill is left just the same as it always was.

  • $\begingroup$ This. You are taking energy from the ball when you decelerate it, then giving it back when you accelerate it in the opposite direction. $\endgroup$ – Time4Tea Aug 26 '18 at 1:04

Ricky Tensor is on the right tracked. However, work is certainly required to make a particle slow to a stop and then speed up in the other direction. If we are assuming a constant force, call it $\vec F = \left\lVert\vec F \right\rVert {-\hat x}$ acting on the particle to change its direction, then the particle will:

1) start at some initial speed $\left\lVert\vec v \right\rVert$ in the +$\hat x$ direction at some intitial position, say $\vec x_0 = 0 \hat x$ or the origin

2) slow to a stop some distance away from its starting positions, say $\vec x_1$

3) now the particle starts to speed up in the -$\hat x$ direction. With the same force being applied the particle will take a distance of $x_1$ to reach speed $\left\lVert\vec v \right\rVert$ in the -$\hat x$ direction

Qualitatively the particle has work done on it that pulls the kinetic energy away until there is no kinetic energy left in the particle;it is now at rest. Then the particle has work done on it that gives it kinetic energy. In fact, the same amount of kinetic energy it starts with since we are assuming the particle ends at the same speed it started with, just in another direction. Ultimately, no net energy has been taken away from or added to the particle so there is no net work done.

Mathematically it can be seen that the work done on the particle in the slowing down or speeding up phase is: $$W =\int_{x_0}^{x_1} \vec F \cdot d\vec x$$

But the net work for both the particle slowing down then speeding back up is: \begin{align} W_{Net} & = \int_{x_0}^{x_1} \vec F \cdot d\vec x\ + \int_{x_1}^{x_0} \vec F \cdot d\vec x \\ & = \int_{x_0}^{x_1} \left\lVert\vec F \right\rVert ({-\hat x} \cdot \hat x)dx + \int_{x_1}^{x_0} \left\lVert\vec F \right\rVert ({-\hat x} \cdot \hat x)dx \\ & = -\left\lVert\vec F \right\rVert\int_{x_0}^{x_1} dx - \left\lVert\vec F \right\rVert\int_{x_1}^{x_0} dx \\ & = -\left\lVert\vec F \right\rVert [x|_{x_0}^{x_1}] + \left\lVert\vec F \right\rVert [x|_{x_0}^{x_1}] \\ & = \left\lVert\vec F \right\rVert \bigl[ -(x_1 - x_0) + (x_1 - x_0)]\bigr] \\ & = 0 \\ \end{align}

Sorry that this was a long winded and probably more that what was needed explanation, but I'm a noob and wanted to practice using Latex syntax. It's been awhile XD


Let the particle of mass $m$ be the system under consideration.
Look at the diagram below which shows the particle initially moving in the $\hat x$ direction at a velocity $\vec v_{\rm initial}= v \,\hat x $ at position $A$ with a constant external force $\vec F = F (-\hat x)$ acting on it in the $(-\hat x$.

The direction of motion being reversed at position $B$.

When the particle reaches position $A$ again it has velocity $\vec v_{\rm final}= v(-\hat x)$.

enter image description here

When going from $A$ to $B$ the work done on the particle by the external force is $\vec F \cdot \vec \Delta \vec x =[F(-\hat x)]\cdot [\Delta x \hat x] = - F\,\Delta x$
The negative sign means that the force is doing negative work on the particle which can be interpreted as the particle doing positive work on the force (ie the kinetic energy of the particle has decreased).

When going from $B$ to $A$ the work done on the particle by the external force is $\vec F \cdot \vec \Delta \vec x =[F(-\hat x)]\cdot [\Delta x (-\hat x)] = + F\,\Delta x$
The positive sign means that the force is doing positive work on the particle (ie the kinetic energy of the particle has increased).

So the total work done by the external force is $(-F\Delta x)+(+F\Delta x) =0$ and this is equal to the change in kinetic energy of the system $\frac 12 mv^2 -\frac 12 mv^2 =0$

An example of such motion although not with a constant force is a particle executing simple harmonic motion.


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